Foundation lesson
Approximating a curve near a point
Close to a point, a smooth curve is described by its value there, its slope and its curvature. Dropping the rest needs a reason: how small the step is, and how fast the dropped terms shrink with it.
Written for this edition, not translated from Einstein.
Why can a curve be replaced, close to one point, by a few simple terms?
What is 2.1 squared, without a calculator? Start from what you know: 2 squared is 4. Picture a square 2 wide growing to 2.1 wide. It gains two strips, each 2 long and 0.1 thick, which add 0.4, and a small corner, 0.1 by 0.1, which adds 0.01. So 2.1 squared is 4.41. The first correction, 0.4, did almost all the work; the second, 0.01, is forty times smaller.
The same recipe works for any smooth curve, close to a point where you know it. Start with its value there. Add its slope times the step Δ: the slope, or first derivative, is the change in output per unit of input. Then add half its curvature times Δ²: the curvature, or second derivative, says how fast the slope itself changes. Further terms carry higher powers of Δ, so they are small only when Δ is:
The value a step delta away equals the value, plus delta times the slope, plus half of delta squared times the curvature, plus further terms.
The Brownian paper does exactly this in §4. It expands the number of particles at x + Δ in powers of the jump Δ and keeps the terms up to Δ², because only very small jumps contribute and each further surviving term is much smaller than the one before. The Δ term averages away, since jumps to the left and to the right are equally likely, and the Δ² term becomes the diffusion coefficient. A real particle does not make independent jumps over arbitrarily short times, so the expansion holds for times long enough for successive jumps to be independent and short enough for each jump to be small.
Worked example: What the dropped term costs
For the square function, the local expansion is exact through the quadratic term.
Keep only the value and the slope term, 4 + 2 × 2 × Δ, and see what dropping Δ² costs as the step grows:
- Δ = 0.1: 4.4 instead of 4.41. The error, 0.01, is about 0.2 per cent.
- Δ = 0.5: 6 instead of 6.25. The error, 0.25, is 4 per cent.
- Δ = 2: 12 instead of 16. The error, 4, is a quarter of the answer.
- Halving the step quarters the error, because the dropped term is Δ².
The square, its two strips and its corner
A square 2 wide grown to 2.1, drawn to scale. The strips are a twentieth of the side and the corner is what the straight-line estimate throws away.
Keeping the square and the strips gives 4.4, which is the straight-line estimate. The whole grown square is 4.41. The difference is the corner, 0.01, and it did about a fiftieth of the work the strips did.
Now the same square at each step the lesson lists, all three at one scale. The corner grows as the square of the step, so it catches up quickly.
What it shows, in words
The first drawing is a square of side 2 and area 4, grown to 2.1. It gains a strip along its right edge and another along its top, each 2 long and 0.1 thick, which together add 0.4, and a small square corner of 0.1 by 0.1, which adds 0.01. Together that is 4.41. Everything is at its true size, so the corner really is the speck it looks like: a four-hundredth of the area.
The second drawing is the same square grown by 0.1, 0.5, 2 in turn, all at one scale. The dropped corners are 0.01, 0.25, 4. At the smallest step the corner is a speck and the straight-line estimate is almost the whole answer; at a step of 2 the corner is a quarter of the square and the estimate has stopped being useful. Halving the step quarters the corner, because the corner is the step squared.
What it does not show: squaring is the case where the series stops of its own accord. A square has one corner and nothing after it, so keeping the value, the strips and the corner is not an approximation at all, it is the whole answer. For a curve that is not a square there are further terms behind the corner, each smaller than the last near the point and each growing faster away from it, and nothing here draws them.
Extension · The step the mass–energy paper takes
Try it: the Lorentz factor at low speed, term by term
The factor 1/√(1 − v²/c²) runs through relativity. It is now called γ; the relativity paper writes it β, with V for the speed of light. The mass–energy paper finds that a body's kinetic energy falls by L(1/√(1 − v²/V²) − 1) when it emits light of energy L, and then expands that bracket. Writing x = v²/c², Newton's binomial series gives:
1. Moderate speeds: partial sums at v/c = 0.6 (x = 0.36)
At sixty percent the speed of light, x = 0.36. The exact value of γ − 1 is 1/√(1 − 0.36) − 1 = 1/0.8 − 1 = 0.25. Add the terms of the series one at a time and the total closes in on it:
| Term added | Its value | Running total | Short of 0.25 |
|---|---|---|---|
| ½ x | 0.180000 | 0.180000 | 0.070000 |
| ⅜ x² | 0.048600 | 0.228600 | 0.021400 |
| ⁵/₁₆ x³ | 0.014580 | 0.243180 | 0.006820 |
| ³⁵/₁₂₈ x⁴ | 0.004593 | 0.247773 | 0.002227 |
Keeping only the first term is the paper's own step: "neglecting quantities of fourth and higher order", it writes the loss of kinetic energy as (L/V²)(v²/2). At v = 0.6c that gives 0.18L where the exact bracket gives 0.25L. The paper needs only low speeds, where the neglected terms are tiny.
2. Low speeds: a computer is also better off with the series
At v/c = 10⁻⁴, about 30 kilometres per second, roughly the Earth's speed round the Sun, x = 10⁻⁸ and γ = 1.000000005. A computer keeps about 16 significant digits, so subtracting 1 from γ throws away the first eight and leaves an answer made mostly of rounding error. The first term of the series, or the bracket rearranged so that nothing close to 1 is subtracted, keeps every digit:
Comparison at v/c = 10⁻⁴ (x = 10⁻⁸)
- First series term (½ x):
- 5.0000000000 × 10−9
- Rearranged, nothing near 1 subtracted:
- 5.0000000375 × 10−9
- Its difference from the first term, relative:
- 7.500000 × 10−9, which the next term predicts: 0.75x
- Direct subtraction, 1/√(1 − x) − 1:
- 5.0000001917 × 10−9
- Its difference from the first term, relative:
- 3.833145 × 10−8, about five times the true difference
Where this lesson stops
An expansion is a claim about size. Say which terms you keep, and why the ones you drop are smaller: here, because they carry higher powers of a small step.
If you came here from a passage, Back returns you to the exact place you left.